Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 946: 58

Answer

$(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=-1$

Work Step by Step

Recall Equation 7, which is: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ $(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=- \dfrac{F_x}{F_y}\dfrac{F_y}{F_x}\dfrac{F_z}{F_y}$ $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ This gives: $(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=-1$
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